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References etc.

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Antti H S Laaksonen 2017-02-25 21:12:39 +02:00
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16
chapter29.tex
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@ -14,7 +14,7 @@ and our task is to calculate its area.
For example, a possible input for the problem is as follows: For example, a possible input for the problem is as follows:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.45] \begin{tikzpicture}[scale=0.44]
\draw[fill] (6,2) circle [radius=0.1]; \draw[fill] (6,2) circle [radius=0.1];
\draw[fill] (5,6) circle [radius=0.1]; \draw[fill] (5,6) circle [radius=0.1];
@ -27,7 +27,7 @@ One way to approach the problem is to divide
the quadrilateral into two triangles by a straight the quadrilateral into two triangles by a straight
line between two opposite vertices: line between two opposite vertices:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.45] \begin{tikzpicture}[scale=0.44]
\draw[fill] (6,2) circle [radius=0.1]; \draw[fill] (6,2) circle [radius=0.1];
\draw[fill] (5,6) circle [radius=0.1]; \draw[fill] (5,6) circle [radius=0.1];
@ -41,7 +41,8 @@ line between two opposite vertices:
After this, it suffices to sum the areas After this, it suffices to sum the areas
of the triangles. of the triangles.
The area of a triangle can be calculated, The area of a triangle can be calculated,
for example, using \key{Heron's formula} for example, using \key{Heron's formula}\footnote{Heron of Alexandria
(c. 10--70) was a Greek mathematician.}
\[ \sqrt{s (s-a) (s-b) (s-c)},\] \[ \sqrt{s (s-a) (s-b) (s-c)},\]
where $a$, $b$ and $c$ are the lengths where $a$, $b$ and $c$ are the lengths
of the triangle's sides and of the triangle's sides and
@ -56,7 +57,7 @@ two arbitrary opposite vertices.
For example, in the following situation, For example, in the following situation,
the division line is outside the quadrilateral: the division line is outside the quadrilateral:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.45] \begin{tikzpicture}[scale=0.44]
\draw[fill] (6,2) circle [radius=0.1]; \draw[fill] (6,2) circle [radius=0.1];
\draw[fill] (3,2) circle [radius=0.1]; \draw[fill] (3,2) circle [radius=0.1];
@ -69,7 +70,7 @@ the division line is outside the quadrilateral:
\end{center} \end{center}
However, another way to draw the line works: However, another way to draw the line works:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.45] \begin{tikzpicture}[scale=0.44]
\draw[fill] (6,2) circle [radius=0.1]; \draw[fill] (6,2) circle [radius=0.1];
\draw[fill] (3,2) circle [radius=0.1]; \draw[fill] (3,2) circle [radius=0.1];
@ -500,7 +501,8 @@ so $b$ is outside the polygon.
\section{Polygon area} \section{Polygon area}
A general formula for calculating the area A general formula for calculating the area
of a polygon is of a polygon\footnote{This formula is sometimes called the
\index{shoelace formula} \key{shoelace formula}.} is
\[\frac{1}{2} |\sum_{i=1}^{n-1} (p_i \times p_{i+1})| = \[\frac{1}{2} |\sum_{i=1}^{n-1} (p_i \times p_{i+1})| =
\frac{1}{2} |\sum_{i=1}^{n-1} (x_i y_{i+1} - x_{i+1} y_i)|, \] \frac{1}{2} |\sum_{i=1}^{n-1} (x_i y_{i+1} - x_{i+1} y_i)|, \]
where the vertices are where the vertices are
@ -571,7 +573,7 @@ along the boundary of the polygon.
\index{Pick's theorem} \index{Pick's theorem}
\key{Pick's theorem} provides another way to calculate \key{Pick's theorem} \cite{pic99} provides another way to calculate
the area of a polygon provided that all vertices the area of a polygon provided that all vertices
of the polygon have integer coordinates. of the polygon have integer coordinates.
According to Pick's theorem, the area of the polygon is According to Pick's theorem, the area of the polygon is

6
chapter30.tex
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@ -270,7 +270,11 @@ we should find the following points:
This is another example of a problem This is another example of a problem
that can be solved in $O(n \log n)$ time that can be solved in $O(n \log n)$ time
using a sweep line algorithm. using a sweep line algorithm\footnote{Besides this approach,
there is also an
$O(n \log n)$ time divide-and-conquer algorithm \cite{sha75}
that divides the points into two sets and recursively
solves the problem for both sets.}.
We go through the points from left to right We go through the points from left to right
and maintain a value $d$: the minimum distance and maintain a value $d$: the minimum distance
between two points seen so far. between two points seen so far.

11
list.tex
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@ -206,6 +206,12 @@
Where to use and how not to use polynomial string hashing. Where to use and how not to use polynomial string hashing.
\emph{Olympiads in Informatics}, 7(1):90--100, 2013. \emph{Olympiads in Informatics}, 7(1):90--100, 2013.
\bibitem{pic99}
G. Pick.
Geometrisches zur Zahlenlehre.
\emph{Sitzungsberichte des deutschen naturwissenschaftlich-medicinischen Vereines
für Böhmen "Lotos" in Prag. (Neue Folge)}, 19:311--319, 1899.
\bibitem{pri57} \bibitem{pri57}
R. C. Prim. R. C. Prim.
Shortest connection networks and some generalizations. Shortest connection networks and some generalizations.
@ -215,6 +221,11 @@
27-Queens Puzzle: Massively Parallel Enumeration and Solution Counting. 27-Queens Puzzle: Massively Parallel Enumeration and Solution Counting.
\url{https://github.com/preusser/q27} \url{https://github.com/preusser/q27}
\bibitem{sha75}
M. I. Shamos and D. Hoey.
Closest-point problems.
\emph{16th Annual Symposium on Foundations of Computer Science}, 151--162, 1975.
\bibitem{sha81} \bibitem{sha81}
M. Sharir. M. Sharir.
A strong-connectivity algorithm and its applications in data flow analysis. A strong-connectivity algorithm and its applications in data flow analysis.